Transition Capacitance of PN Junction Diodes

Transition Capacitance (C_T)

• When a diode is reverse biased, the width of the depletion region increases. So there are more positive and negative charges present in the depletion region.

• Due to this, the p‒region and n‒region act like the plates of capacitor while the depletion region acts like dielectric.

• Thus there exists a capacitance at the p‒n junction called transition capacitance, junction capacitance, space charge capacitance, barrier capacitance or depletion region capacitance. It is denoted as C_T.

• Mathematically it is given by the expression,

Ст = εΑ / W

      ……... (2.13.1)

where

ε = Permittivity of semiconductor

= ε₀ε_r

ε₀ = 1 / 36π×10⁹ = 8.849×10^‒12 F/m

ε_r = Relative permittivity of semiconductor = 16 for Ge, 12 for Si

A = Area of cross section

W = Width of depletion region

• As the reverse biased applied to the diode increases, the width of the depletion region (W) increases. Thus the transition capacitance C_T decreases.

• In short, the capacitance can be controlled by the applied voltage. The variation of C_T with respect to the applied reverse bias voltage is shown in Fig. 2.13.2.

• As reverse voltage is negative, graph is shown in the second quadrant. For a particular diode shown, C_T varies from 80 pF to less than 5 pF as V_R changes from 2 V to 15 V.

1. Derivation of Expression for Transition Capacitance

• Consider the p‒n junction whose p side is lightly doped than n side.

• Such an unequally doped p‒n junction is shown in Fig. 2.13.3, which is reverse biased.

• The depletion region penetrates more on lightly doped side hence its width is more on p side.

• Practically it can be assumed that concentration of impurity on both sides is such that the entire depletion region is on p side only.

• Let N_A = Concentration of acceptor impurity on p side.

• The relation between potential V and charge density N_A is given by Poisson's equation as,

 d²V / dx² = qN_A / ε

     ….. x=Distance measured from junction

  ε = ε₀ ε_r

 = Permittivity of semiconductor

where

 ε₀ = 8.854 × 10^‒12 F/m

 ε_r = 16 for Ge and 12 for Si

• Integrating above equation with respect x,

 ...(2.13.2)

• But dV/dx is the electric field E over the region x = 0 to x = W over which depletion region exists.

E = qN_Ax / ε

       ……. q = Charge on each electron ...(2.13.3)

• Integrating equation (2.13.2)

         ...(2.13.4)

• At x=W, V = V_B = Barrier potential which is the difference between junction potential V_J and applied reverse bias V i.e. V_B = V_J− V

         ...(2.13.5)

• V must be taken as negative as reverse biased. Hence as V becomes more negative, V_B = V_J − (− V) = V_J + V increases.

• Using in equation (2.13.4),

VB = qN_A/ε . W²/2

W ∝ √V_B

    ...(2.13.6)

The width of depletion region increases as applied reverse bias voltage increases

• If A is the area of cross‒section of junction, W is the width then net charge Q in the distance W is,

 Q = N_A × Volume × q = N_A A W q       ...(2.13.7)

• Differentiating equation (2.13.4) with respect to V,

...(2.13.8)

• Differentiating equation (2.13.7) with respect to V,

dQ / dV = N_A A (dW/dV) q

         ...(2.13.9)

• Using equation (2.13.8) in (2.13.9) and dQ/dV = C_T

Ст = εA/W =  Transition capacitance

  ………….. (2.13.10)

• Thus C_T = 1/W and hence as reverse bias increases, the width of the depletion region W increases and hence transition capacitance C_T decreases.

Ex. 2.13.1: A germanium diode has a contact potential of 0.2 V and concentration of acceptor impurity is 3×10²⁰/m³. Area of cross‒section of junction is 1 mm² and ε_r = 16. Calculate width of depletion region and transition capacitance for reverse bias of a) 0.1 V and b) 10 V (Assume charge on each electron as 1.6 × 10^‒19C)

Solution:

 V₁ = 0.2 V, N_A = 3 × 10²⁰/m³, A = 1 mm², ε_r = 16

a) V = ‒0.1 V

       ……… Negative as reverse bias

 V_B = V_J‒ V = 0.2 − (− 0.1) = 0.3 V

 V_B = qN_A/ε . W²/2

 i.e.

 0.3 = (1.6 × 10^‒19 × 3 × 10²⁰) / (8.854 × 10^‒12 × 16) . W²/2

 W = 1.33 μm,

 С_т = εA/W = (ε×1×10^‒19) / (1.33×10^‒6) = 106.45 pF

b) V = ‒10V

hence

V_B = V_J ‒ V = 0.2 ‒ (‒10) = 10.2 V

 10.2 = [ 1.6 × 10^‒19 × 3 × 10²⁰ ] / [8.854 × 10^‒12 × 16] × [W²/2]

i.e.

W = 7.75 μm

Ст = εA

= ( 8854 × 10^‒12 × 16 × 1 × 10^‒6 ) / (7.75 × 10^‒6)

= 18.25 pF

Thus as reverse bias voltage increases, width of depletion region increases while transition capacitance decreases.

Review Questions

1. Explain the concept of transition capacitance of p‒n junction diode.

2. Derive the expression for the capacitance of p‒n junction diode.